Question

Find the values of $k$ for which the quadratic equation $9x^2-3kx+k=0$ has equal roots.

Answer 1

We know that $D=0$ then the equation has equal and real roots

$D=b^2-4ac=0$

$a=9,b=-3k,c=k$

Subsitute the values we get,

$(-3k{)}^2-4\times 9\times k=0$

$\Rightarrow 9k^2-36k=0$

$\Rightarrow 9k^2=36k$

$\Rightarrow 9k(k-4)=0$

$\therefore k=0\text{or}k=4$